Question Number 162726 by LEKOUMA last updated on 31/Dec/21
$$\left.\mathrm{1}\right)\:{Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{\left(\mathrm{sin}\:{x}\right)^{{n}} },\:\:\left({m},\:{n}\in\: {N}\right) \\ $$$$\left.\mathrm{2}\right)\:{f}'\left({a}\right)\:{e}\mathrm{xiste},\:{calculate} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}\left[{f}\left({a}+\frac{{a}}{{x}}\right)−{f}\left({a}−\frac{\beta}{{x}}\right)\right],\: \\ $$$$\left(\alpha,\:\beta\:\in\: {R}\right) \\ $$
Answered by mr W last updated on 01/Jan/22
$$\left(\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{\left(\mathrm{sin}\:{x}\right)^{{n}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{{x}^{{m}} }×\left(\frac{{x}}{\mathrm{sin}\:{x}}\right)^{{n}} ×{x}^{{m}−{n}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{{m}−{n}} =\begin{cases}{\mathrm{1}\:{if}\:{m}={n}}\\{\mathrm{0}\:{if}\:{m}>{n}}\\{\infty\:{if}\:{m}<{n}}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}\left[{f}\left({a}+\frac{\alpha}{{x}}\right)−{f}\left({a}\right)−{f}\left({a}−\frac{\beta}{{x}}\right)+{f}\left({a}\right)\right]\: \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left[\frac{{f}\left({a}+\frac{\alpha}{{x}}\right)−{f}\left({a}\right)}{\frac{\mathrm{1}}{{x}}}−\frac{{f}\left({a}−\frac{\beta}{{x}}\right)−{f}\left({a}\right)}{\frac{\mathrm{1}}{{x}}}\right]\: \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left[\alpha×\frac{{f}\left({a}+\frac{\alpha}{{x}}\right)−{f}\left({a}\right)}{\frac{\alpha}{{x}}}+\beta×\frac{{f}\left({a}−\frac{\beta}{{x}}\right)−{f}\left({a}\right)}{−\frac{\beta}{{x}}}\right]\: \\ $$$$=\underset{{h},{k}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\alpha×\frac{{f}\left({a}+{h}\right)−{f}\left({a}\right)}{{h}}+\beta×\frac{{f}\left({a}+{k}\right)−{f}\left({a}\right)}{{k}}\right]\: \\ $$$$=\alpha{f}'\left({a}\right)+\beta{f}'\left({a}\right) \\ $$$$=\left(\alpha+\beta\right){f}'\left({a}\right) \\ $$
Commented by Ar Brandon last updated on 01/Jan/22
👏👏👏First comment of the year (GMT+1)
Happy New year, Sir !
Happy New year, Sir !
Commented by mr W last updated on 01/Jan/22
$${thanks}! \\ $$$${the}\:{same}\:{to}\:{you}\:{and}\:{all}\:{others}! \\ $$
Commented by Ar Brandon last updated on 01/Jan/22
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$